00:01
So, in the question we have to find the directional derivative of the function at the given point in the direction of the vector v.
00:07
The given function is f x, y is equal to 1 plus 2 x under root y and the point that is given 3, 4 and the directional vector v is equal to 4, minus 3.
00:25
So, the formula for the directional derivative is directional derivative is equal to gradient into unit direction vector.
00:34
F x, y is given that so, first of all we have to find the grad f.
00:43
The formula for grad f is i dev f over dev x plus j dev f over dev y and plus k dev f over dev z, but we have only two terms.
00:56
Then the formula is in the place of f we have to put 1 plus 2 x under root y that is i cap dev 1 plus 2 x under root y upon dev x plus j cap dev 1 plus 2 x under root y upon dev y.
01:29
Now, we have to differentiate it then we get 1 plus 2 under root y i cap plus 1 plus 2 x upon under root 2 y j cap 2 and 2 cancel.
01:56
We have to find grad f at point 3, 4 p is equal to 3, 4.
02:09
So, in the place of x and y we have to put 3 and 4...